Problem: A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(10t,6t^{4}\right)$ for time $t\geq 0$. At $t=0$, the particle is at the point $(-8,-3)$. What is the particle's position at $t=2$ ? $($
Solution: To find the particle's position at $t=2$, we need to find its horizontal displacement $\Delta x$ and its vertical displacement $\Delta y$, and add those to its initial position $(-8,-3)$ : $\text{Position at }t=2\text{: }(-8+\Delta x,-3+\Delta y)$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between $t=0$ and $t=2$ : $\Delta x=\int_{0}^{2} 10t\,dt=20$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between $t=0$ and $t=2$ : $\Delta y=\int_{0}^{2} 6t^{4}\,dt=\dfrac{192}{5}$ Now we can find the particle's position: $\begin{aligned} &\phantom{=}(-8+\Delta x,-3+\Delta y) \\\\ &=\left(-8+20,-3+\dfrac{192}{5}\right) \\\\ &=(12,35.4) \end{aligned}$ In conclusion, particle's position at $t=2$ is $(12,35.4)$.